Question

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent, then the value of $\lambda$ is.

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is D

$2$

Explanation for the correct option:

Step 1: Find the point of intersection of lines given by the pair of straight lines.

In the question, a pair of straight lines $x^2+2xy-35y^2-4x+44y-12=0$ is given.

We know that, if a pair of straight lines $a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0$, then the point of intersection of lines is given by $\left(\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2}\right)$.

On comparing both the equations we get $a=1,b=-35,c=-12,g=-2,h=1$ and $f=22$.

So, the point of intersection of lines given by $x^2+2xy-35y^2-4x+44y-12=0$ can be given as follows:

$$\begin{gathered} \left(\frac{1·22-(-35)(-2)}{\left(1\right)(-35)-{\left(1\right)}^2},\frac{(-2)\left(1\right)-\left(1\right)\left(22\right)}{\left(1\right)(-35)-{\left(1\right)}^2}\right)=\left(\frac{22-70}{-36},\frac{-2-22}{-36}\right) \\ \Rightarrow \left(\frac{1·22-(-35)(-2)}{\left(1\right)(-35)-{\left(1\right)}^2},\frac{(-2)\left(1\right)-\left(1\right)\left(22\right)}{\left(1\right)(-35)-{\left(1\right)}^2}\right)=\left(\frac{4}{3},\frac{2}{3}\right) \end{gathered}$$

Therefore, the point of intersection of lines given by the equation of pair of straight lines is $\left(\frac{4}{3},\frac{2}{3}\right)$.

Step 2: Find the value of $\lambda$.

Since the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent and the point of intersection of lines given by the equation of pair of straight lines is $\left(\frac{4}{3},\frac{2}{3}\right)$.

So, the line $5x+\lambda y-8=0$ also satisfy $\left(\frac{4}{3},\frac{2}{3}\right)$.

That is,

$$\begin{gathered} 5\left(\frac{4}{3}\right)+\lambda \left(\frac{2}{3}\right)-8=0 \\ \Rightarrow 20+2\lambda -24=0 \\ \Rightarrow 2\lambda -4=0 \\ \Rightarrow \lambda =2 \end{gathered}$$

Therefore, the value of $\lambda$ is $2$.

Hence, option (D) is the correct answer.